Question: Factor the quadratic expression completely. $6x^2-13x+6=$
Explanation: Since the terms in the expression do not share a common monomial factor and the coefficient on the leading $x^2$ term is not $1$, let's factor by grouping. The expression ${6}x^2{-13}x{+6}$ is in the form ${A}x^2+{B}x+{C}$. First, we need to find two integers ${a}$ and ${b}$ such that: $\begin{cases} &{a}+{b}={B}={-13} \\\\ &{ab}={A}{C}= ({6})({6})=36 \end{cases}$ We find that ${a}={-4}$ and ${b}={-9}$ satisfy these conditions, since ${-4}+({-9})={-13}$ and $({-4})({-9})=36$. Next, we can use these values to rewrite the $x$ -term and factor by grouping. $\begin{aligned} 6x^2-13x+6&=6x^2{-4}x{-9}x+6 \\\\ &=2x(3x-2)-3(3x-2) \\\\ &=(3x-2)(2x-3) \end{aligned}$ In conclusion, $6x^2-13x+6=(3x-2)(2x-3)$